THE RATIONAL NUMBERSbase-ten system to fractions was given by Simon Stevin (1548–1620),who introduced to 16th-century Europe a method of writing decimal fractions. The decimal that

$Page 1: THE RATIONAL NUMBERSbase-ten system to fractions was given by Simon Stevin (1548–1620),who introduced to 16th-century Europe a method of writing decimal fractions. The decimal that$
CHAPTER

2

39

CHAPTERTABLE OF CONTENTS

2-1 Rational Numbers

2-2 Simplifying RationalExpressions

2-3 Multiplying and DividingRational Expressions

2-4 Adding and SubtractingRational Expressions

2-5 Ratio and Proportion

2-6 Complex RationalExpressions

2-7 Solving Rational Equations

2-8 Solving Rational Inequalities

Chapter Summary

Vocabulary

Review Exercises

Cumulative Review

THERATIONALNUMBERS

When a divided by b is not an integer, the quotientis a fraction.The Babylonians, who used a number sys-tem based on 60, expressed the quotients:

20 � 8 as instead of

21 � 8 as instead of

Note that this is similar to saying that 20 hoursdivided by 8 is 2 hours, 30 minutes and that 21 hoursdivided by 5 is 2 hours, 37 minutes, 30 seconds.This notation was also used by Leonardo of Pisa(1175–1250), also known as Fibonacci.

The base-ten number system used throughout theworld today comes from both Hindu and Arabic math-ematicians. One of the earliest applications of the base-ten system to fractions was given by Simon Stevin(1548–1620), who introduced to 16th-century Europea method of writing decimal fractions. The decimal that we write as 3.147 was written by Stevin as

3 � 1 � 4 � 7 � or as 3�

1�

4�

7�

. John Napier(1550–1617) later brought the decimal point into com-mon usage.

2582 1 37

60 1 303,600

2122 1 30

60

14411C02.pgs 8/12/08 1:47 PM Page 39

When persons travel to another country, one of the first things that they learn isthe monetary system. In the United States, the dollar is the basic unit, but most purchases require the use of a fractional part of a dollar. We know

that a penny is $0.01 or of a dollar, that a nickel is $0.05 or of a

dollar, and a dime is $0.10 or of a dollar. Fractions are common in our everyday life as a part of a dollar when we make a purchase, as a part of a poundwhen we purchase a cut of meat, or as a part of a cup of flour when we arebaking.

In our study of mathematics, we have worked with numbers that are notintegers. For example, 15 minutes is or of an hour, 8 inches is or of a foot,and 8 ounces is or of a pound. These fractions are numbers in the set ofrational numbers.

For every rational number that is not equal to zero, there is a multiplicative

inverse or reciprocal such that 5 1. Note that . If the non-zeronumerator of a fraction is equal to the denominator, then the fraction is equalto 1.

EXAMPLE 1

Write the multiplicative inverse of each of the following rational numbers:

Answers

a.

b.

c. 5

Note that in b, the reciprocal of a negative number is a negative number.

15

825 5 28

5258

43

34

ab ? ba 5 ab

abab ? ba

ba

ab

12

816

23

812

14

1560

10100 5 1

10

5100 5 1

201

100

2-1 RATIONAL NUMBERS

40 The Rational Numbers

DEFINITION

A rational number is a number of the form where a and b are integers and b � 0.

ab

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Decimal Values of Rational Numbers

The rational number is equivalent to a � b. When a fraction or a division suchas 25 � 100 is entered into a calculator, the decimal value is displayed. Toexpress the quotient as a fraction, select Frac from the menu. This can bedone in two ways.

ENTER: 25 100 ENTER: 25 100

DISPLAY: DISPLAY:

When a calculator is used to evaluate a fraction such as or 8 � 12, the dec-imal value is shown as .6666666667. The calculator has rounded the value to tendecimal places, the nearest ten-billionth.The true value of , or , is an infinitelyrepeating decimal that can be written as . The line over the 6 means that thedigit 6 repeats infinitely. Other examples of infinitely repeating decimals are:

5 0.22222222 . . . 5

5 0.142857142857 . . . 5

5 0.1212121212 . . . 5

5 0.1666666666 . . . 5

Every rational number is either a finite decimal or an infinitely repeatingdecimal. Because a finite decimal such as 0.25 can be thought of as having an infinitely repeating 0 and can be written as , the following statement is true:

� A number is a rational number if and only if it can be written as an infinitelyrepeating decimal.

0.250

160.16

4330.12

170.142857

290.2

0.6

23

812

812

F r a c1 / 4

2 5 / 1 0 0

A n s F r a c1 / 4

. 2 52 5 / 1 0 0

ENTER1MATH

ENTER1MATH�ENTER�

MATH

ab

Rational Numbers 41

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EXAMPLE 2

Find the common fractional equivalent of .

Solution Let x 5 5 0.18181818 . . .

How to Proceed(1) Multiply the value of x by 100 to write a

number in which the decimal point follows the first pair of repeating digits:

(2) Subtract the value of x from both sides of this equation:

(3) Solve the resulting equation for x and simplify the fraction:

Check The solution can be checked on a calculator.

ENTER: 2 11

DISPLAY:

Answer

EXAMPLE 3

Express as a common fraction.

Solution: Let x 5 0.12484848. . .

How to Proceed(1) Multiply the value of x by the power of 10 that

makes the decimal point follow the first set of repeating digits. Since we want to move the decimal point 4 places, multiply by 104 5 10,000:

(2) Multiply the value of x by the power of 10 that makes the decimal point follow the digits that do not repeat. Since we want to move the decimal point 2 places, multiply by 102 5 100:

(3) Subtract the equation in step 2 from the equation in step 1:

(4) Solve for x and reduce the fraction to lowest terms:

Answer 103825

0.1248

211

. 1 8 1 8 1 8 1 8 1 82 / 1 1

ENTER�

0.18

0.18


100x 5 18.181818 . . .

x 5 1899 5 2

11

99x 5 182x 5 20.181818c

100x 5 18.181818c

10,000x 5 1,248.4848. . .

100x 5 12.4848. . .

x 51,2369,900 5 103

825

9,900x 5 1,2362100x 5 212.4848c

10,000x 5 1,248.4848c

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Writing About Mathematics

1. a. Why is a coin that is worth 25 cents called a quarter?

b. Why is the number of minutes in a quarter of an hour different from the number ofcents in a quarter of a dollar?

2. Explain the difference between the additive inverse and the multiplicative inverse.

Developing SkillsIn 3–7, write the reciprocal (multiplicative inverse) of each given number.

3. 4. 5. 6. 8 7. 1

In 8–12, write each rational number as a repeating decimal.

8. 9. 10. 11. 12.

In 13–22, write each decimal as a common fraction.

13. 0.125 14. 15. 16. 17.

18. 19. 20. 21. 22. 0.15900.1360.570.830.156

0.1080.360.20.6

78

215

57

29

16

227

712

38

Exercises

Rational Numbers 43

Procedure

To convert an infinitely repeating decimal to a common fraction:

1. Write the equation: x 5 decimal value.

2. Multiply both sides of the equation in step 1 by 10m, where m is the numberof places to the right of the decimal point following the first set of repeatingdigits.

3. Multiply both sides of the equation in step 1 by 10n, where n is the numberof places to the right of the decimal point following the non-repeating digits.(If there are no non-repeating digits, then let n 5 0.)

4. Subtract the equation in step 3 from the equation in step 2.

5. Solve the resulting equation for x, and simplify the fraction completely.

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A rational number is the quotient of two integers. A rational expression is thequotient of two polynomials. Each of the following fractions is a rational expres-sion:

Division by 0 is not defined. Therefore, each of these rational expressionshas no meaning when the denominator is zero. For instance:

• has no meaning when x 5 0.

• has no meaning when a 5 0 or when b 5 0.

• has no meaning when y 5 2 or when y 5 3.

EXAMPLE 1

For what value or values of a is the fraction undefined?

Solution A fraction is undefined or has no meaning when a factor of the denominatoris equal to 0.

How to Proceed

(1) Factor the denominator: 3a2 2 4a 1 1 5 (3a 2 1)(a 2 1)

(2) Set each factor equal to 0: 3a 2 1 5 0 a 2 1 5 0

(3) Solve each equation for a: 3a 5 1 a 5 1

a 5

Answer The fraction is undefined when a 5 and when a 5 1.

� If and are rational numbers with b � 0 and d � 0, then

and

For example: and .We can write a rational expression in simplest form by finding common fac-

tors in the numerator and denominators, as shown above.

312 5 3 3 1

3 3 4 5 33 3 1

4 5 1 3 14 5 1

434 3 1

3 5 312

acbd 5 a

b ? cd 5 ad ?

cb

ab ? cd 5 ac

bd

cd

ab

13

13

2a 2 53a2 2 4a 1 1

y 2 2y2 2 5y 1 6 5

y 2 2(y 2 2)(y 2 3)

a2 2 14ab

x 1 52x

y 2 2

y22 5y 1 6

a2 2 14ab

x 1 52x

ab7

34

2-2 SIMPLIFYING RATIONAL EXPRESSIONS


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EXAMPLE 2

Simplify: Answers

a.

b.

c.

Note: We must eliminate any value of the variable or variables for which thedenominator of the given rational expression is zero.

The rational expressions , , and in the example shown above are in simplest form because there is no factor of the numerator that is also a factorof the denominator except 1 and 21. We say that the fractions have beenreduced to lowest terms.

When the numerator or denominator of a rational expression is a mono-mial, each number or variable is a factor of the monomial. When the numeratoror denominator of a rational expression is a polynomial with more than oneterm, we must factor the polynomial. Once both the numerator and denomina-tor of the fraction are factored, we can reduce the fraction by identifying factorsin the numerator that are also factors in the denominator.

In the example given above, we wrote:

We can simplify this process by canceling the common factor in the numer-ator and denominator.

(y � 2, 3)

Note that canceling (y 2 2) in the numerator and denominator of the frac-tion given above is the equivalent of dividing (y 2 2) by (y 2 2).When any num-ber or algebraic expression that is not equal to 0 is divided by itself, the quotientis 1.

(y 2 2)1

(y 2 2)1

(y 2 3) 5 1y 2 3

y 2 2y2 2 5y 1 6 5

y 2 2(y 2 2)(y 2 3) 5

y 2 2y 2 2 ? 1

y 2 3 5 1 ? 1y 2 3 5

1y 2 3 (y 2 2, 3)

1y 2 3

3ba 2 1

x3

5y 2 2

(y 2 2)(y 2 3) 5y 2 2y 2 2 ? 1

y 2 3 5 1 ? 1y 2 3 5

1y 2 3 (y 2 2, 3)

y 2 2y2 2 5y 1 6

5 aa ? 3b

a 2 1 5 1 ? 3ba 2 1 5 3b

a 2 1 (a 2 0, 1)3aba(a 2 1)

5 22 ? x3 5 1 ? x3 5 x

32x6

Simplifying Rational Expressions 45

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EXAMPLE 3

Write in lowest terms.

Solution METHOD 1 METHOD 2

Answer (x � 0)

Factors That Are Opposites

The binomials (a 2 2) and (2 2 a) are opposites or additive inverses. If wechange the order of the terms in the binomial (2 2 a), we can write:

(2 2 a) 5 (2a 1 2) 5 21(a 2 2)

x 2 4x

5 x 2 4x

5 1 ? x 2 4x

5 x 2 4x5 3

3 ? x 2 4x

3x 2 123x 5

31(x 2 4)

31x

3x 2 123x 5

3(x 2 4)3x

3x 2 123x


Procedure

To reduce a fraction to lowest terms:

METHOD 11. Factor completely both the numerator and the denominator.

2. Determine the greatest common factor of the numerator and the denominator.

3. Express the given fraction as the product of two fractions, one of which hasas its numerator and its denominator the greatest common factor deter-mined in step 2.

4. Write the fraction whose numerator and denominator are the greatestcommon factor as 1 and use the multiplication property of 1.

METHOD 21. Factor both the numerator and the denominator.

2. Divide both the numerator and the denominator by their greatest commonfactor by canceling the common factor.

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We can use this factored form of (2 2 a) to reduce the rational expressionto lowest terms.

EXAMPLE 4

Simplify the expression:

Solution METHOD 1 METHOD 2

Answer (x � 4)


1. Abby said that can be reduced to lowest terms by canceling 3x so that the result is .Do you agree with Abby? Explain why or why not.

2. Does 5 1 for all real values of a? Justify your answer.

Developing SkillsIn 3–10, list the values of the variables for which the rational expression is undefined.

3. 4. 5. 6.

7. 8. 9. 10. 5x3 2 5x2 2 6x

4c2c2 2 2c

b 1 3b2 1 b 2 6

2a2a 2 7

x 2 5x 1 5

a 1 2ab

22d6c

5a2

3a

2a 2 32a 2 3

14

3x3x 1 4

Exercises

22x 2 4

5 22x 2 45 22

x 2 4

522(x 2 4)

1

(x 2 4)(x 2 4)1

5(x 2 4)(x 2 4) ? 22

(x 2 4)

52(21)(x 2 4)

(x 2 4)(x 2 4)52(21)(x 2 4)

(x 2 4)(x 2 4)

8 2 2xx2 2 8x 1 16 5

2(4 2 x)(x 2 4)(x 2 4)

8 2 2xx2 2 8x 1 16 5

2(4 2 x)(x 2 4)(x 2 4)

8 2 2xx2 2 8x 1 16

2 2 aa2 2 4 5

21(a 2 2)(a 1 2)(a 2 2) 5

21(a 2 2)1

(a 1 2)(a 2 2)1

521

(a 1 2) 5 21

a 1 2 (a 2 2, 22)

2 2 aa2 2 4

Simplifying Rational Expressions 47

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In 11–30, write each rational expression in simplest form and list the values of the variables forwhich the fraction is undefined.

11. 12. 13. 14.

15. 16. 17. 18.

19. 20. 21. 22.

23. 24. 25. 26.

27. 28. 29. 30.

Multiplying Rational Expressions

We know that and that . In general, the product of two

rational numbers and is for b � 0 and d � 0.This same rule holds for the product of two rational expressions:

� The product of two rational expressions is a fraction whose numerator is theproduct of the given numerators and whose denominator is the product ofthe given denominators.

For example:

(a � 0)

This product can be reduced to lowest terms by dividing numerator and denom-inator by the common factor, 4.

We could have canceled the factor 4 before we multiplied, as shown below.

Note that a is not a common factor of the numerator and denominator becauseit is one term of the factor (a 1 5), not a factor of the numerator.

3(a 1 5)4a 3

12a 5

3(a 1 5)41a 3 12

3

a 59(a 1 5)

a2

3(a 1 5)4a ? 12

a 536(a 1 5)

4a2 536

9(a 1 5)

41a2 5

9(a 1 5)a2

3(a 1 5)4a ? 12

a 536(a 1 5)

4a2

ab ? cd 5 ac

bdcd

ab

34 3 3

2 5 98

23 3 4

5 5 815

2-3 MULTIPLYING AND DIVIDING RATIONAL EXPRESSIONS

5(1 2 b) 1 15b2 2 16

4 2 2(x 2 1)x2 2 6x 1 9

3 2 (b 1 1)4 2 b2

a3 2 a2 2 a 1 1a2 2 2a 1 1

27 1 7a21a2 2 21

5y2 2 20y2 1 4y 1 4

x2 2 7x 1 12x2 1 2x 2 15

4a2 2 164a 1 8

2a 1 103a 1 15

3xy9xy 1 6x2y3

8c2

8c2 1 16c10d

15d 2 20d2

8ab 2 4b2

6ab9y2 1 3y

6y28a 1 16

12a9cd2

12c4d2

14b4

21b312xy2

3x2y5a2b10a

610


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EXAMPLE 1

a. Find the product of in simplest form.

b. For what values of the variable are the given fractions and the productundefined?

Solution a. METHOD 1

How to Proceed

(1) Multiply the numerators of the fractions and the denominators of the fractions:

(2) Factor the numerator and the denominator. Note thatthe factors of (5b 1 15) are 5(b 1 3) and the factors of (3b2 1 9b) are 3b(b 1 3).Reduce the resulting fraction to lowest terms:

12b2

5b 1 15 ? b2 2 93b2

1 9b

Multiplying and Dividing Rational Expressions 49

Procedure

To multiply fractions:

METHOD 11. Multiply the numerators of the given fractions and the denominators of the

given fractions.

2. Reduce the resulting fraction, if possible, to lowest terms.

METHOD 21. Factor any polynomial that is not a monomial.

2. Cancel any factors that are common to a numerator and a denominator.

3. Multiply the resulting numerators and the resulting denominators to writethe product in lowest terms.

Answer54b(b 2 3)5(b 1 3)

5 1 ? 4b(b 2 3)5(b 1 3)

53b(b 1 3)3b(b 1 3) ?

4b(b 2 3)5(b 1 3)

512b2(b 1 3)(b 2 3)15b(b 1 3)(b 1 3)

12b2

5b 1 15 ? b2 2 93b2 1 9b 5

12b2(b2 2 9)(5b 1 15)(3b2 1 9b)

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METHOD 2

How to Proceed

(1) Factor each binomial term completely:

(2) Cancel any factors that are common to a numerator and a denominator:

(3) Multiply the remaining factors:

b. The given fractions and their product are undefined when b 5 0 andwhen b 5 23. Answer

Dividing Rational Expressions

We can divide two rational numbers or two rational expressions by changing thedivision to a related multiplication. Let and be two rational numbers or ratio-nal expressions with b � 0, c � 0, d � 0. Since is a non-zero rational number orrational expression, there exists a multiplicative inverse such that .

We have just derived the following procedure.

For example:

(x � 21, 0)

This product is in simplest form because the numerator and denominator haveno common factors.

Recall that a can be written as and therefore, if a � 0, the reciprocal of ais . For example:

(b � 1)2b 2 25 4 (b 2 1) 5

2(b 2 1)5 ? 1

b 2 1 52(b 2 1)

1

5 ? 1b 2 1

1

5 25

1a

a1

5(x 1 5)(x 1 1)

10x2

x 1 52x2 4 5

x 1 1 5 x 1 52x2 ? x 1 1

5

ab 4

cd 5

abcd

5

abcd

? dcdc

5

ab ? dccd ? dc

5

ab ? dc

1 5 ab ? d

c

cd ? dc 5 1d

c

cd

cd

ab


Answer54b(b 2 3)5(b 1 3)

512

4b2b

5(b 1 3) ? (b 1 3)1

(b 2 3)31b1(b 1 3)

1

12b2

5b 1 15 ? b2 2 93b2 1 9b 5

12b2

5(b 1 3) ? (b 1 3)(b 2 3)3b(b 1 3)

Procedure

To divide two rational numbers or rational expressions, multiply thedividend by the reciprocal of the divisor.

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EXAMPLE 2

Divide and simplify:

Solution How to Proceed

(1) Use the reciprocal of the divisor to write the division as a multiplication:

(2) Factor each polynomial:

(3) Cancel any factors thatare common to the numerator and denominator:

(4) Multiply the remaining factors:

Answer (a � 0, 5)

EXAMPLE 3

Perform the indicated operations and write the answer in simplest form:

Solution Recall that multiplications and divisions are performed in the order in whichthey occur from left to right.

Answer5 2a5 (a 2 23, 0)

5 62

5 ? a31

53a

1

a 1 31

? 2(a 1 3)1

5a1

? a3

3aa 1 3 4

5a2a 1 6 ? 3a 5

3aa 1 3 ? 2(a 1 3)

5a ? a3

3aa 1 3 4 5a

2a 1 6 ? 3a

3(a 1 2)a2

a2 2 3a 2 105a 4 a2 2 5a

15

Multiplying and Dividing Rational Expressions 51

53(a 1 2)

a2

5(a 2 5)

1(a 1 2)

51a ? 15

3

a(a 2 5)1

5(a 2 5)(a 1 2)

5a ? 15a(a 2 5)

a2 2 3a 2 105a 4 a2 2 5a

15 5 a2 2 3a 2 105a ? 15

a2 2 5a

14411C02.pgs 8/12/08 1:47 PM Page 51


1. Joshua wanted to write this division in simplest form: . He began by cancel-

ing (x 2 2) in the numerator and denominator and wrote following:

Is Joshua’s answer correct? Justify your answer.

2. Gabriel wrote . Is Gabriel’s solution correct? Justifyyour answer.

Developing SkillsIn 3–12, multiply and express each product in simplest form. In each case, list any values of the vari-ables for which the fractions are not defined.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

In 13–24, divide and express each quotient in simplest form. In each case, list any values of the vari-ables for which the fractions are not defined.

13. 14.

15. 16.

17. 18.

19. 20.

21. 22.

23. 24. (a2 2 1) 4 2a 1 2a(2x 1 7) 4 1

2x2 1 5x 2 7

a2 1 8a 1 154a 4 (a 1 3)4b 1 12

b 4 (b 1 3)

w2 2 w5w 4 w2 2 1

5c2 2 6c 1 9

5c 2 15 4 c 2 35

6y2 2 3y3y 4

4y2 2 12

x 2 23x 4 4x 2 8

9

a2

8a 4 3a4

6b5c 4 3b

10c

12a 4 6

4a34 4 9

20

6 2 2xx2 2 9 ? 15 1 5x

4x2a 1 4

6a ? 3a2

a2 1 2a

a2 2 5a 1 43a 1 6 ? 2a 1 4

a2 2 167y 1 21

7y ? 3y2 2 9

a2 2 1003a ? a2

2a 2 20b 1 1

4 ? 125b 1 5

3a5 ? 10

9a4y5x ? x

8y

57a ? 3a

2023 3 3

4

12x5x 1 10 4 4

5 5 12x 4 4(5x 1 10) 4 5 5 3x

x 1 2

3x 2 2

1

44(x 2 2)

1

7 5 31 4 4

7 5 31 3 7

4 5 214

3x 2 2 4

4(x 2 2)7

Exercises


14411C02.pgs 8/12/08 1:47 PM Page 52

In 25–30, perform the indicated operations and write the result in simplest form. In each case, listany values of the variables for which the fractions are not defined.

25. 26.

27. 28.

29. 30.

We know that and that . Therefore:

We can also write:

In general, the sum of any two fractions with a common denominator is afraction whose numerator is the sum of the numerators and whose denomina-tor is the common denominator.

(a � 21)

In order to add two fractions that do not have a common denominator, weneed to change one or both of the fractions to equivalent fractions that do havea common denominator. That common denominator can be the product of thegiven denominators.

For example, to add , we change each fraction to an equivalent fractionwhose denominator is 35 by multiplying each fraction by a fraction equal to 1.

5 1235

5 735 1 5

35

15 1 1

7 5 15 3 7

7 1 17 3 5

5

15 1 1

7

3aa 1 1 1 a

a 1 1 5 4aa 1 1

5 3x 2 3x (x 2 0)

5 (x 1 2 1 2x 2 5) A 1x B

x 1 2x 1 2x 2 5

x 5 (x 1 2) A 1x B 1 (2x 2 5) A 1

x B

5 57

5 5 A 17 B

5 (3 1 2) A 17 B

37 1 2

7 5 3 A 17 B 1 2 A 1

7 B27 5 2 A 1

7 B37 5 3 A 1

7 B2-4 ADDING AND SUBTRACTING RATIONAL EXPRESSIONS

x2 2 3x 1 24x ? 12x2

x2 2 2x 4 x 2 1x(3b)2 4 3b

b 1 2 ? 2b 1 4b

(x2 2 2x 1 1) 4 x 2 13 ? x 1 4

3x2a

a 1 2 ? a2 2 44a2 4 a 2 2

a

3xx 2 1 ? x

2 2 1x 4 x 1 1

335 3 5

9 4 43

Adding and Subtracting Rational Expressions 53

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To add the fractions and , we need to find a denominator that is a mul-tiple of both 2x and of y. One possibility is their product, 2xy. Multiply each frac-tion by a fraction equal to 1 so that the denominator of each fraction will be 2xy:

and

(x � 0, y � 0)

The least common denominator (LCD) is often smaller than the product of the two denominators. It is the least common multiple (LCM) of thedenominators, that is, the product of all factors of one or both of the denominators.

For example, to add , first find the factors of each denomina-tor.The least common denominator is the product of all of the factors of the firstdenominator times all factors of the second that are not factors of the first.Thenmultiply each fraction by a fraction equal to 1 so that the denominator of eachfraction will be equal to the LCD.

Factors of 2a 1 2: 2 � (a 1 1)Factors of a2 2 1: (a 1 1) � (a 2 1)

LCD: 2 � (a 1 1) � (a 2 1)

and

Since this sum has a common factor in the numerator and denominator, itcan be reduced to lowest terms.

(a � 21, a � 1)

Any polynomial can be written as a rational expression with a denominatorof 1. To add a polynomial to a rational expression, write the polynomial as anequivalent rational expression.

For example, to write the sum b 1 3 1 as a single fraction, multiply (b 1 3) by 1 in the form .

5 2b2 1 6b 1 12b (b 2 0)

5 2b2 1 6b2b 1 1

2b

(b 1 3) 1 12b 5 b 1 3

1 A 2b2b B 1 1

2b

2b2b

12b

a 1 12(a 1 1)(a 2 1) 5 a 1 1

1

2(a 1 1)(a 2 1)1

5 12(a 2 1)

12a 1 2 1 1

a2 2 1 5 a 2 1 1 22(a 1 1)(a 2 1) 5 a 1 1

2(a 1 1)(a 2 1)

5 22(a 1 1)(a 2 1)5 a 2 1

2(a 1 1)(a 2 1)

1a2 2 1 5 1

(a 1 1)(a 2 1) ? 221

2a 1 2 5 12(a 1 1) ? a 2 1

a 2 1

12a 1 2 1 1

a2 2 1

52x 1

3y 5

5y2xy 1

6x2xy 5

5y 1 6x2xy

3y 5 3

y ? 2x2x 5 6x

2xy5

2x 55

2x ? yy 5

5y2xy

3y

52x


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EXAMPLE 1

Write the difference as a single fraction in lowest terms.

Solution How to Proceed(1) Find the LCD of the

fractions:

(2) Write each fraction as an equivalent fraction with a denominator equal to the LCD:

(3) Subtract:

(4) Simplify:

(5) Reduce to lowest terms:

Answer (x � 23, 1, 3)

EXAMPLE 2

Simplify:

Solution STEP 1. Rewrite each expression in parentheses as a single fraction.

and

STEP 2. Multiply.

5 x 1 1

5 x 1 11

5 ¢ (x 1 1)(x 2 1) 1

x1

≤ ¢ x1

x 2 11

≤A x2 2 1

x B A xx 2 1 B 5 Q(x 1 1)(x 2 1)

x R A xx 2 1 B

5 xx 2 1

5 x 2 1 1 1x 2 15 x2 2 1

x

5 x 2 1x 2 1 1 1

x 2 15 x2

x 2 1x

1 1 1x 2 1 5 1 A x 2 1

x 2 1 B 1 1x 2 1x 2 1

x 5 x A xx B 2 1

x

Ax 2 1x B A1 1 1

x 2 1 B

9(x 2 3)(x 1 3)

xx2 2 4x 1 3 2 x

x2 1 2x 2 3

Adding and Subtracting Rational Expressions 55

x2 2 4x 1 3 5 (x 2 3) � (x 2 1)

x2 1 2x 2 3 5 (x 2 1) � (x 1 3)

LCD 5 (x 2 3) � (x 2 1) � (x 1 3)

5 9(x 2 3)(x 1 3)

5 9x 2 9(x 2 3)(x 2 1)(x 1 3)

5x2 1 3x 2 (x2 2 6x 1 9)

(x 2 3)(x 2 1)(x 1 3)

xx2 2 4x 1 3 2 x 2 3

x2 1 2x 2 3

5 x2 2 6x 1 9(x 2 3)(x 2 1)(x 1 3)

x 2 3x2 1 2x 2 3 5 x 2 3

(x 1 3)(x 2 1) ? x 2 3x 2 3

5 x2 1 3x(x 2 3)(x 2 1)(x 1 3)

xx2 2 4x 1 3 5 x

(x 2 3)(x 2 1) ? x 1 3x 1 3

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STEP 1. Multiply using the distributive property.

STEP 2. Add the fractions. The least common denominator is x(x 2 1).

STEP 3. Simplify.

Answer x 1 1 (x � 0, 1)


1. Ashley said that for all values of a except a 5 23. Do you agree with

Ashley? Explain why or why not.

2. Matthew said that when b � 0, d � 0. Do you agree with Matthew? Justifyyour answer.

Developing SkillsIn 3–20, perform the indicated additions or subtractions and write the result in simplest form. Ineach case, list any values of the variables for which the fractions are not defined.

3. 4. 5.

6. 7. 8.

9. 10. 11.

12. 13. 14.

15. 16. 17.

18. 19. 20. 1x 1 1

x 2 2 2 2x2 2 2x

2a2 2 4 2 1

a2 1 2a1

a2 2 a 2 6 2 12a2 2 7a 1 3

12 2 x 1 2

x 2 2b

b 2 1 2 12 2 2b

3x 1 2 1 x 2 2

x

1x 1 1a 2 3

2a5 2 12y

3 1 2x

a 1 13a 1 3

22x 1 3

6x 2 x 2 24x

a 1 55a 2 a 2 8

8ay 1 2

2 12y 2 3

3a 2 1

5 2 a 1 14

x7 1 x

32x2 1 1

5x 2 7x2 2 15x

x3 1 2x

3

ab 1 c

d 5 ad 1 bcbd

(a 1 2)(a 2 1)(a 1 3)(a 2 1) 5

a 1 2a 1 3

Exercises

x3 2 xx(x 2 1) 5

x1(x 2 1

1)(x 1 1)

x1(x 2 1

1) 5 x 1 1

5 x3 2 xx(x 2 1)

5 x3 2 x2 1 x2 2 x 1 1 2 1x(x 2 1)

5 x3 2 x2

x(x 2 1) 1 x2

x(x 2 1) 2 x 2 1x(x 2 1) 2 1

x(x 2 1)

x 1x

x 2 1 21x 2

1x(x 2 1) 5 xQx(x 2 1)

x(x 2 1) R 1x

(x 2 1) A xx B 2

1x A x 2 1

x 2 1 B 21

x(x 2 1)

5 x 1 xx 2 1 2 1

x 2 1x(x 2 1)

Ax 2 1x B A1 1 1

x 2 1 B 5 x A 1 1 1x 2 1 B 2 1

x A1 1 1x 2 1 B

AlternativeSolution


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Applying SkillsIn 21–24, the length and width of a rectangle are expressed in terms of a variable.

a. Express each perimeter in terms of the variable.

b. Express each area in terms of the variable.

21. l 5 2x and w 5

22. l 5 3x 1 3 and w 5

23. l 5 and w 5

24. l 5 and w 5

We often want to compare two quantities that use the same unit. For example,in a given class of 25 students, there are 11 students who are boys. We can saythat of the students are boys or that the ratio of students who are boys to allstudents in the class is 11 : 25.

A ratio, like a fraction, can be simplified by dividing each term by the samenon-zero number. A ratio is in simplest form when the terms of the ratio areintegers that have no common factor other than 1.

For example, to write the ratio of 3 inches to 1 foot, we must first write eachmeasure in terms of the same unit and then divide each term of the ratio by acommon factor.

In lowest terms, the ratio of 3 inches to 1 foot is 1 : 4.An equivalent ratio can also be written by multiplying each term of the ratio

by the same non-zero number. For example, 4 : 7 5 4(2) : 7(2) 5 8 : 14.In general, for x � 0:

a : b 5 ax : bx

3 inches1 foot 5 3 inches

1 foot 3 1 foot12 inches 5 3

12 5 14

1125

2-5 RATIO AND PROPORTION

xx 1 2

xx 1 1

3x 2 1

xx 2 1

13

1x

Ratio and Proportion 57

DEFINITION

A ratio is the comparison of two numbers by division. The ratio of a to b canbe written as or as a : b when b � 0.a

b

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EXAMPLE 1

The length of a rectangle is 1 yard and the width is 2 feet. What is the ratio oflength to width of this rectangle?

Solution The ratio must be in terms of the same measure.

Answer The ratio of length to width is 3 : 2.

EXAMPLE 2

The ratio of the length of one of the congruent sides of an isosceles triangle tothe length of the base is 5 : 2. If the perimeter of the triangle is 42.0 centimeters,what is the length of each side?

Solution Let AB and BC be the lengths of the congruent sides of isosceles �ABC andAC be the length of the base.

AB : AC 5 5 : 2 5 5x : 2x

Therefore, AB 5 5x,

BC 5 5x,

and AC 5 2x.

AB 1 BC 1 AC 5 Perimeter

5x 1 5x 1 2x 5 42

12x 5 42

x 5 3.5 cm

Check AB 1 BC 1 AC 5 17.5 1 17.5 1 7.0 5 42.0 cm ✔

Answer The sides measure 17.5, 17.5, and 7.0 centimeters.

5 7.0 cm5 17.5 cm

AC 5 2(3.5)AB 5 BC 5 5(3.5)

1 yd2 ft 3 3 ft

1 yd 5 32


B

A C

5x5x

2x

14411C02.pgs 8/12/08 1:47 PM Page 58

Proportion

An equation that states that two ratios are equal is called a proportion. Forexample, 3 : 12 5 1 : 4 is a proportion. This proportion can also be written as

.In general, if b � 0 and d � 0, then a : b 5 c : d or are proportions.The

first and last terms, a and d, are called the extremes of the proportion and thesecond and third terms, b and c, are the means of the proportion.

If we multiply both sides of the proportion by the product of the second andlast terms, we can prove a useful relationship among the terms of a proportion.

� In any proportion, the product of the means is equal to the product of theextremes.

EXAMPLE 3

In the junior class, there are 24 more girls than boys. The ratio of girls to boys is5 : 4. How many girls and how many boys are there in the junior class?

Solution How to Proceed

(1) Use the fact that the number of girls is 24 more than the number of boys to represent the number of girls and of boys in terms of x:

(2) Write a proportion. Set the ratio of the number of boys to the number of girls, in terms of x, equal to the given ratio:

(3) Use the fact that the product of the means is equal to the product of the extremes. Solve the equation:

da 5 bc

b1d¢a

b1

≤ 5 bd1 ¢ c

d1

≤bd A a

b B 5 bd A cd B

ab 5 c

d

ab 5 c

d

312 5 1

4

Ratio and Proportion 59

Let x 5 the number of boys,

x 1 24 5 the number of girls.

5 120

x 1 24 5 96 1 24

x 5 96

5x 5 4x 1 96

5x 5 4(x 1 24)

x 1 24x 5 5

4

14411C02.pgs 8/12/08 1:47 PM Page 59

(1) Use the given ratio to represent the number of boys and the number of girlsin terms of x:

(2) Use the fact that the number of girls is 24 more than the number of boys to write an equation. Solve the equation for x:

(3) Use the value of x to find the number of girls and the number of boys:

Answer There are 120 girls and 96 boys in the junior class.


1. If , then is ? Justify your answer.

2. If , then is ? Justify your answer.

Developing SkillsIn 3–10, write each ratio in simplest form.

3. 12 : 8 4. 21 : 14 5. 3 : 18 6. 15 : 75

7. 6a : 9a, a � 0 8. 9. 10. , x � 0

In 11–19, solve each proportion for the variable.

11. 12. 13.

14. 15. 16.

17. 18. 19.

Applying Skills

20. The ratio of the length to the width of a rectangle is 5 : 4. The perimeter of the rectangle is72 inches. What are the dimensions of the rectangle?

21. The ratio of the length to the width of a rectangle is 7 : 3. The area of the rectangle is 336square centimeters. What are the dimensions of the rectangle?

22. The basketball team has played 21 games. The ratio of wins to losses is 5 : 2. How manygames has the team won?

x5 5 x 1 4

x 1 13x 2 2

x 5 3x 1 2

4x 2 83 5 8

x 2 3

2x 2 1 5 x 1 2

23x 1 3

16 5 2x 1 110

y 1 3y 1 8 5 6

15

a 2 22a 5 5

14x 2 1

15 5 25

x8 5 6

24

10x35x

2472

1827

a 1 bb 5 c 1 d

dab 5 c

d

ac 5 b

dab 5 c

d

Exercises

AlternativeSolution


Let 5x 5 the number of girls

4x 5 the number of boys

5x 5 24 1 4x

x 5 24

5x 5 5(24) 5 120 girls

4x 5 4(24) 5 96 boys

14411C02.pgs 8/12/08 1:47 PM Page 60

23. In the chess club, the ratio of boys to girls is 6 : 5. There are 3 more boys than girls in theclub. How many members are in the club?

24. Every year, Javier makes a total contribution of $125 to two local charities. The two dona-tions are in the ratio of 3 : 2. What contribution does Javier make to each charity?

25. A cookie recipe uses flour and sugar in the ratio of 9 : 4. If Nicholas uses 1 cup of sugar,how much flour should he use?

26. The directions on a bottle of cleaning solution suggest that the solution be diluted withwater. The ratio of solution to water is 1 : 7. How many cups of solution and how many cupsof water should Christopher use to make 2 gallons (32 cups) of the mixture?

A complex fraction is a fraction whose numerator, denominator, or both containfractions. Some examples of complex fractions are:

A complex fraction can be simplified by multiplying by a fraction equal to1; that is, by a fraction whose non-zero numerator and denominator are equal.The numerator and denominator of this fraction should be a common multipleof the denominators of the fractional terms. For example:

A complex fraction can also be simplified by dividing the numerator by thedenominator.

A complex rational expression has a rational expression in the numerator,the denominator, or both. For example, the following are complex rationalexpressions.

1 1 1b 2

2b2

1b

x 1 31x

1aa

23418

5

84 1 3

418

5

11418

5 114 4 1

8 5 114 3 8

1 5 884 5 22

527

5 51 4 2

7 5 5 3 72 5 35

2

132 5 1

3 4 21 5 1

3 3 12 5 1

6

23418

52 1 3

418

3 88 5 16 1 6

1 5 22527

5 527

3 77 5 35

2

132 5

132 3 3

3 5 16

23418

527

132

2-6 COMPLEX RATIONAL EXPRESSIONS

Complex Rational Expressions 61

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Like complex fractions, complex rational expressions can be simplified bymultiplying numerator and denominator by a common multiple of the denomi-nators in the fractional terms.

Alternatively, a complex rational expression can also be simplified by divid-ing the numerator by the denominator. Choose the method that is easier foreach given expression.

EXAMPLE 1

Express in simplest form.

Solution METHOD 1

Multiply numerator and denominator of the fraction by the least commonmultiple of the denominators of the fractional terms. The least common multi-ple of a and 4a2 is 4a2.

2a3

4a2

52a3

4a2

? 4a2

4a2 5 8a3

2a3

4a2

5 b 1 2b (b 2 0, 1)

5(b 1 2)(b 2 1)

1

b(b 2 1)1

(x 2 0)

5(b 1 2)(b 2 1)

b(b 2 1)5 x2 1 3x(a 2 0)

5 b2 1 b 2 2b2 2 b5 x2 1 3x

15 1a2

1 1 1b 2 2

b2

1 2 1b

51 1 1

b 2 2b2

1 2 1b

? b2

b2x 1 3

1x

5 x 1 31x

? xx

1aa 5

1aa ? aa


Procedure

To simplify a complex fraction:METHOD 1Multiply by , where m is the least common multiple of the denominators of the fractional terms.METHOD 2Multiply the numerator by the reciprocal of the denominator of the fraction.

mm

14411C02.pgs 8/12/08 1:47 PM Page 62

METHOD 2

Write the fraction as the numerator divided by the denominator. Change thedivision to multiplication by the reciprocal of the divisor.

Answer (a � 0)

EXAMPLE 2

Simplify:

SolutionMETHOD 1 METHOD 2

The least common multiple of 10, 2,and 25 is 50.

Answer (b � 25, 5)


1. For what values of a is undefined? Explain your answer.

2. Bebe said that since each of the denominators in the complex fraction is a non-zero

constant, the fraction is defined for all values of d. Do you agree with Bebe? Explain why orwhy not.

d4 1 3

5

2 2 d2

2

1 2 1a

1 2 1a2

A1 2 1a B 4 A1 2 1

a2 B 5

Exercises

52(b 2 5)

5 52(b 2 5)

55(b 1 5)

2(b 1 5)(b 2 5)

5 5b 1 252b2

2 50

b10 1 1

2b2

25 2 15

b10 1 1

2b2

25 2 1 ? 50

50

b10 1 1

2b2

25 2 1

8a3

2a3

4a2

5 2a 4 3

4a2 5 2a ? 4a2

3 5 2a1

? 4a2a

3 5 8a3

Complex Rational Expressions 63

5 52(b 2 5)

5 b 1 51

102

? 255

(b 1 5)1

(b 2 5)

5 b 1 510 4 b2 2 25

25

5

b 1 510

b2 2 2525

b10 1 1

2b2

25 2 15

b10 1 5

10b2

25 2 2525

14411C02.pgs 8/12/08 1:47 PM Page 63

Developing SkillsIn 3–20, simplify each complex rational expression. In each case, list any values of the variables forwhich the fractions are not defined.

3. 4. 5.

6. 7. 8.

9. 10. 11.

12. 13. 14.

15. 16. 17.

18. 19. 20.

In 21–24, simplify each expression. In each case, list any values of the variables for which the frac-tions are not defined.

21. 22.

23. 24.

An equation in which there is a fraction in one or more terms can be solved indifferent ways. However, in each case, the end result must be an equivalentequation in which the variable is equal to a constant.

2-7 SOLVING RATIONAL EQUATIONS

A6 1 12b B 4 A3b 2 12

b B 1 b2 2 bA 3

a 1 5a2 B 4 A 10

a 1 6 B 1 34

a4 1 7a

86a2

5 2 3a2

101 3

a3

2x 21 1 1

xx 1 1

5 2 45a2

3a 2 1

1a 2 1

1 2 1a

1 1 1y 2 6

y2

1 1 11y 1 24

y2

1 1 3b 1 2

b2

1 2 1b2

3 2 9x

x 2 8 1 15x

a 2 49a

a 2 9 1 14a

12x 1 1

3x1x2

4y 2 1y

y 2 12

y 1 12

2y 1 1

3 2 3b

b 2 1

b 2 1b

1b 2 1

2 2 2d

1d 2 1

1 1 1a

a 1 1

16 1 1

413 1 1

2

2x1

2x

78134

254

334


14411C02.pgs 8/12/08 1:47 PM Page 64

For example, solve: . This is an equation with rational coeffi-

cients and could be written as . There are three possible meth-ods of solving this equation.

METHOD 1

Work with the fractions as given.Combine terms containing the variable on one side of the equation and constants on the other. Find common denominators to add or subtract fractions.

METHOD 2

Rewrite the equation without fractions.Multiply both sides of the equation by the least common denominator. In this case, the LCD is 4.

METHOD 3

Rewrite each side of the equation as a ratio. Use the fact that the product of the means is equal to the product of the extremes.

These same procedures can also be used for a rational equation, an equa-tion in which the variable appears in one or more denominators.

14x 2 2 5 1

2x 1 12

x4 2 2 5 x

2 1 12

Solving Rational Equations 65

x 5 52(24) 5 220

2 5 210

2x4 5 5

2

x4 2 2x

4 5 12 1 4

2

x4 2 x

2 5 12 1 2

x4 2 2 5 x

2 1 12

x 5 210

2x 5 10

x 2 8 5 2x 1 2

4 A x4 2 2 B 5 4 A x

2 1 12 B

x4 2 2 5 x

2 1 12

x 5 210

2x 5 220

4x 1 4 5 2x 2 16

4(x 1 1) 5 2(x 2 8)

x 2 84 5 x 1 1

2

x4 2 8

4 5 x2 1 1

2

x4 2 2 5 x

2 1 12

14411C02.pgs 8/12/08 1:47 PM Page 65

EXAMPLE 1

Solve the equation .

Solution Use Method 2.

How to Proceed Check(1) Write the equation:

(2) Multiply both sides of the equation bythe least common denominator, 6x:

(3) Simplify:

(4) Solve for x:

Answer x 5

When solving a rational equation, it is important to check that the fractionalterms are defined for each root, or solution, of the equation. Any solution forwhich the equation is undefined is called an extraneous root.

EXAMPLE 2

Solve for a:

Solution Since this equation is a proportion, we can use that the product of the meansis equal to the product of the extremes. Check the roots.

2a(a 2 2) 5 a(a 1 2)

2a2 2 4a 5 a2 1 2a

a2 2 6a 5 0

a(a 2 6) 5 0

a 5 0 a 2 6 5 0

a 5 6

Answer a 5 6

a 2 2a 5 a 1 2

2a

a 2 2a 5 a 1 2

2a

38

x 5 38

8x 5 3

6 1 2x 5 9 2 6x

6xx 1 6x

3 5 18x2x 2 6x

6x A 1x 1 1

3 B 5 6x A 32x 2 1 B

1x 1 1

3 5 32x 2 1

1x 1 1

3 5 32x 2 1


✔3 5 3

93 5? 3

83 1 1

3 5? 4 2 1

A1 3 83 B 1 1

3 5? A 3 3 43 B 2 1

3

2Q38R

2 15?1

38

1 13

1x 1 1

3 5 32x 2 1

Check: a 5 0

Each side of theequation is undefinedfor a 5 0. Therefore,a 5 0 is not a root.

0 2 20 5? 0 1 2

2(0)

a 2 2a 5 a 1 2

2a

Check: a 5 6

✔23 5 2

3

46 5? 8

12

6 2 26 5? 6 1 2

2(6)

a 2 2a 5 a 1 2

2a

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EXAMPLE 3

Solve:

Solution Use Method 2 to rewrite the equation without fractions. The least commondenominator is x(x 1 2).

x 2 5 5 0 x 1 2 5 0

x 5 5 x 5 22

Check the roots:

Check: x 5 5 Check: x 5 2 2

✔

Since x 5 5 leads to a true statement,5 is a root of the equation.

Answer 5 is the only root of .

It is important to check each root obtained by multiplying both members ofthe given equation by an expression that contains the variable. The derivedequation may not be equivalent to the given equation.

xx 1 2 5 3

x 1 4x(x 1 2)

2535 5 25

35

2535 5? 21

35 1 435

57 3 5

5 5? 35 3 77 1 4

35

55 1 2 5? 35 1 4

5(5 1 2)

xx 1 2 5 3

x 1 4x(x 1 2)

(x 2 5)(x 1 2) 5 0

x2 2 3x 2 10 5 0

x2 5 3x 1 6 1 4

x(x 1 2)1 a x

x 1 21

b 5 x1(x 1 2)a3

x1

b 1 x1(x 1 2)

1 a 4x1(x 1 2)

1

bx(x 1 2) A x

x 1 2 B 5 x(x 1 2) A 3x B 1 x(x 1 2) A 4

x(x 1 2) Bx

x 1 2 5 3x 1 4

x(x 1 2)

xx 1 2 5 3

x 1 4x(x 1 2)


Since x 5 22 leads to a statementthat is undefined, 22 is not a rootof the equation.

220 5? 23

2 1 422(0)

2222 1 2 5? 3

22 1 422(22 1 2)

xx 1 2 5 3

x 1 4x(x 1 2)

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EXAMPLE 4

Solve and check: 0.2y 1 4 5 5 2 0.05y

Solution The coefficients 0.2 and 0.05 are decimal fractions (“2 tenths” and “5 hun-dredths”). The denominator of 0.2 is 10 and the denominator of 0.05 is 100.Therefore, the least common denominator is 100.

Answer y 5 4

EXAMPLE 5

On his way home from college, Daniel trav-eled 15 miles on local roads and 90 miles onthe highway. On the highway, he traveled 30miles per hour faster than on local roads. Ifthe trip took 2 hours, what was Daniel’s rateof speed on each part of the trip?

Solution Use to represent the time for each part of the trip.

Let x 5 rate on local roads. Then 5 time on local roads.

Let x 1 30 5 rate on the highway. Then 5 time on the highway.

The total time is 2 hours.

Recall that a trinomial can be factored by writing the middle term as the sum oftwo terms whose product is the product of the first and last term.

2x2(2450) 5 2900x2

Find the factors of this product whose sum is the middle term 245x.

15x 1 (260x) 5 245x

0 5 2x2 2 45x 2 450

15x 1 450 1 90x 5 2x2 1 60x

15(x 1 30) 1 90x 5 2x2 1 60x

x(x 1 30) A 15x B 1 x(x 1 30) A 90

x 1 30 B 5 2x(x 1 30)

15x 1 90

x 1 30 5 2

90x 1 30

15x

distancerate 5 time

y 5 4

25y 5 100

20y 1 400 5 500 2 5y

100(0.2y 1 4) 5 100(5 2 0.05y)

0.2y 1 4 5 5 2 0.05y


Check

✔4.8 5 4.8

0.8 1 4 5? 5 2 0.2

0.2(4) 1 4 5? 5 2 0.05(4)

0.2y 1 4 5 5 2 0.05y

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Write the trinomial with four terms, using this pair of terms in place of 245x,and factor the trinomial.

0 5 2x2 2 45x 2 450

0 5 2x2 1 15x – 60x 2 450

0 5 x(2x 1 15) 2 30(2x 1 15)

0 5 (2x 1 15)(x 2 30)

2x 1 15 5 0 x 2 30 5 0

2x 5 215 x 5 30

x 5

Reject the negative root since a rate cannot be a negative number. Use x 5 30.

On local roads, Daniel’s rate is 30 miles per hour and his time is hour.

On the highway, Daniel’s rate is 30 1 30 or 60 miles per hour. His time is hours. His total time is hours.

Answer Daniel drove 30 mph on local roads and 60 mph on the highway.


1. Samantha said that the equation in Example 2 could be solved by multiplyingboth sides of the equation by 2a. Would Samantha’s solution be the same as the solutionobtained in Example 2? Explain why or why not.

2. Brianna said that is a rational equation but is not. Do youagree with Brianna? Explain why or why not.

Developing SkillsIn 3–20, solve each equation and check.

3. 4. 5.

6. 7. 8. x 2 0.05x 5 19

9. 0.4x 1 8 5 0.5x 10. 0.2a 5 0.05a 1 3 11. 1.2b 2 3 5 7 2 0.05b

12. 13. 14.

15. 16. 17.

18. 19. 20. 43b 2 2 2 7

3b 1 2 5 19b2

2 44

y 1 2 5 1 2 8y(y 1 2)

a 2 14 5 8

a 1 3

1 5 5x 1 3 1 5

(x 1 2)(x 1 3)x2 5 3

2x 1 118 2 4b 5 10

3a 1 1

2 5 115a

72x 2 3 5 4

xx

x 1 5 5 23

2x3 1 1 5 3x

4x5 2 x

10 5 7

x 1 25 5 x 2 2

334x 5 14 2 x1

4a 1 8 5 12a

x 2 23 5 x 1 2

53

x 2 2 5 5x 1 2

a 2 2a 5 a 1 2

2a

Exercises

12 1 3

2 5 42 5 290

60 5 32

1530 5 1

2

2152


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Applying Skills

21. Last week, the ratio of the number of hours that Joseph worked to the number of hours thatNicole worked was 2 : 3. This week Joseph worked 4 hours more than last week and Nicoleworked twice as many hours as last week. This week the ratio of the hours Joseph worked tothe number of hours Nicole worked is 1 : 2. How many hours did each person work eachweek?

22. Anthony rode his bicycle to his friend’s house, a distance of 1 mile. Then his friend’s motherdrove them to school, a distance of 12 miles. His friend’s mother drove at a rate that is 25miles per hour faster than Anthony rides his bike. If it took Anthony of an hour to get toschool, at what average rate does he ride his bicycle? (Use for each part ofthe trip to school.)

23. Amanda drove 40 miles. Then she increased her rate of speed by 10 miles per hour anddrove another 40 miles to reach her destination. If the trip took hours, at what rate didAmanda drive?

24. Last week, Emily paid $8.25 for x pounds of apples. This week she paid $9.50 for (x 1 1)pounds of apples. The price per pound was the same each week. How many pounds ofapples did Emily buy each week and what was the price per pound? (Use

for each week.)

Inequalities are usually solved with the same procedures that are used to solveequations. For example, we can solve this equation and this inequality by usingthe same steps.

All steps leading to the solution of this equation and this inequality are thesame, but special care must be used when multiplying or dividing an inequality.Note that when we multiplied the inequality by 24, a positive number, the orderof the inequality remained unchanged. In the last step, when we divided theinequality by 22, a negative number, the order of the inequality was reversed.

When we solve an inequality that has a variable expression in the denomi-nator by multiplying both sides of the inequality by the variable expression, wemust consider two possibilities: the variable represents a positive number or the

x . 23x 5 23

22x , 622x 5 6

6x 1 3 , 8x 1 96x 1 3 5 8x 1 9

24 A 14x B 1 24 A 1

8 B , 24 A 13x B 1 24 A 3

8 B24 A 14x B 1 24 A 1

8 B 5 24 A 13x B 1 24 A 3

8 B14x 1 1

8 , 13x 1 3

814x 1 1

8 5 13x 1 3

8

2-8 SOLVING RATIONAL INEQUALITIES

total costnumber of pounds 5 cost per pound

145

distancerate 5 time

35


14411C02.pgs 8/12/08 1:47 PM Page 70

variable represents a negative number. (The expression cannot equal zero asthat would make the fraction undefined.)

For example, to solve , we multiply both sides of theinequality by x 1 1. When x 1 1 is positive, the order of the inequality remainsunchanged. When x 1 1 is negative, the order of the inequality is reversed.

If x 1 1 � 0, then x � –1. If x 1 1 � 0, then x � –1.

Therefore, x � 21 and x � 4 or There are no values of x such 21 � x � 4. that x � 21 and x � 4.

The solution set of the equation is {x : �1 � x � 4}.

An alternative method of solving this inequality is to use the correspondingequation.

STEP 1. Solve the corresponding equation.

STEP 2. Find any values of x for which the equation is undefined.

Terms and are undefined when x 5 21.

STEP 3. To find the solutions of the corresponding inequality, divide the num-ber line into three intervals using the solution to the equality, 4, and thevalue of x for which the equation is undefined, 21.

�3 �2 �1 0 1 2 3 4 5 6

9x 1 1

xx 1 1

x 5 4

22x 5 28

2x 1 2 2 x 5 3x 1 3 2 9

(x 1 1)(2) 2 (x 1 1)1 x

x 1 11

5 (x 1 1)(3) 2 (x 1 1)1 9

x 1 11

2 2 xx 1 1 5 3 2 9

x 1 1

2 2 xx 1 1 . 3 2 9

x 1 1

x . 4x , 4

22x22 . 28

2222x22 , 28

22

9 2 x , x 1 19 2 x . x 1 1

(x 1 1)1 9 2 x

x 1 11

, (x 1 1)(1)(x 1 1)1 9 2 x

x 1 11

. (x 1 1)(1)

9x 1 1 2 x

x 1 1 . 3 2 29x 1 1 2 x

x 1 1 . 3 2 2

2 2 xx 1 1 . 3 2 9

x 1 12 2 xx 1 1 . 3 2 9

x 1 1

2 2 xx 1 1 . 3 2 9

x 1 1

Solving Rational Inequalities 71

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Choose a value from each section and substitute that value into the inequality.

Let x 5 22: Let x 5 0: Let x 5 5:

✘ ✔ ✘

The inequality is true for values in the interval 21 � x � 4 and false for allother values.

EXAMPLE 1

Solve for a:

Solution METHOD 1

Multiply both sides of the equation by a. The sense of the inequality willremain unchanged when a � 0 and will be reversed when a � 0.

Let a � 0: Let a � 0:

a � 0 and a � 4 → a � 4 a � 0 and a � 4 → a � 0

The solution set is {a : a � 0 or a � 4}.

METHOD 2

Solve the corresponding equation for a.

a 5 4

2a 5 8

2a 2 3 5 5

a A2 2 3a B 5 a A 5

a B2 2 3

a 5 5a

a , 4a . 4

2a , 82a . 8

2a 2 3 , 52a 2 3 . 5

a A2 2 3a B , a A 5

a Ba A2 2 3a B . a A 5

a B

2 2 3a . 5

a

76 6

962 . 260 6 12

2 2 56 .

?3 2 9

62 2 (0) .?

3 2 (9)2 2 (2) .?

3 2 (29)

2 2 55 1 1 .

?3 2 9

5 1 12 2 00 1 1 .

?3 2 9

0 1 12 2 2222 1 1 .

?3 2 9

22 1 1

2 2 xx 1 1 . 3 2 9

x 1 12 2 xx 1 1 . 3 2 9

x 1 12 2 xx 1 1 . 3 2 9

x 1 1

�3 �2 �1 0 1 2 3 4 5 6


14411C02.pgs 8/12/08 1:47 PM Page 72

Partition the number line using the solution to the equation, a 5 4, and thevalue of a for which the equation is undefined, a 5 0. Check in the inequalitya representative value of a from each interval of the graph.

Let a 5 21: Let a 5 1: Let a 5 5:

✔ ✘ ✔

Answer {a : a � 0 or a � 4}


1. When the equation is solved for b, the solutions are 21 and 3. Explain whythe number line must be separated into five segments by the numbers 22, 21, 0, and 3 inorder to check the solution set of the inequality .

2. What is the solution set of ? Justify your answer.

Developing SkillsIn 3–14, solve and check each inequality.

3. 4. 5.

6. 7. 8.

9. 10. 11.

12. 13. 14. 3 2 9a 1 1 . 5 2 1

a 1 1x

x 1 5 2 1x 1 5 . 42

x 1 3x , 10

4xx 2 4 1 2 , 2

x 2 43 2 2a 1 1 , 55 2 7

y , 2 1 5y

72x 2 2

x . 32

a 1 14 2 2 . 11 2 a

62 2 d

7 . d 2 25

3b 2 48 , 4b 2 3

4y 2 3

5 ,y 1 2

10a4 . a

2 1 6

�x�x , 0

2 2 3b . 5

b 1 2

2 2 3b 5 5

b 1 2

Exercises

75 . 121 6 55 . 25

105 2 3

5 .?

12 2 3 .?

52 1 3 .?

25

2 2 35 .

? 552 2 3

1 .? 5

12 2 321 .

? 521

2 2 3a . 5

a2 2 3a . 5

a2 2 3a . 5

a

�3 �2 �1 0 1 2 3 4 5 6

Solving Rational Inequalities 73

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CHAPTER SUMMARY

A rational number is an element of the set of numbers of the form whena and b are integers and b � 0. For every rational number that is not equal to

zero there is a multiplicative inverse or reciprocal such that .A number is a rational number if and only if it can be written as an infinitely

repeating decimal.A rational expression is the quotient of two polynomials. A rational expres-

sion has no meaning when the denominator is zero.A rational expression is in simplest form or reduced to lowest terms when

there is no factor of the numerator that is also a factor of the denominatorexcept 1 and 21.

If and are two rational numbers with b � 0 and d � 0:

(c � 0)

A ratio is the comparison of two numbers by division.The ratio of a to b canbe written as or as a : b when b � 0.

An equation that states that two ratios are equal is called a proportion. Inthe proportion a : b 5 c : d or , the first and last terms, a and d, are calledthe extremes, and the second and third terms, b and c, are the means.

In any proportion, the product of the means is equal to the product of theextremes.

A complex rational expression has a rational expression in the numerator,the denominator or both. Complex rational expressions can be simplified bymultiplying numerator and denominator by a common multiple of the denomi-nators in the numerator and denominator.

A rational equation, an equation in which the variable appears in one ormore denominators, can be simplified by multiplying both members by a com-mon multiple of the denominators.

When a rational inequality is multiplied by the least common multiple ofthe denominators, two cases must be considered: when the least common multi-ple is positive and when the least common multiple is negative.

VOCABULARY

2-1 Rational number • Multiplicative inverse • Reciprocal

2-2 Rational expression • Simplest form • Lowest terms

2-4 Least common denominator (LCD) • Least common multiple (LCM)

2-5 Ratio • Ratio in simplest form • Proportion • Extremes • Means

2-6 Complex fraction • Complex rational expression

2-7 Rational equation • Extraneous root

ab 5 c

d

ab

ab 1 c

d 5 ab ? dd 1 c

d ? bb 5 adbd 1 cb

bd 5 ad 1 cbbd

ab 4 c

d 5 ab ? dc 5 ad

bcab ? cd 5 ac

bd

cd

ab

ab ? ba 5 1b

a

ab

ab


14411C02.pgs 8/12/08 1:47 PM Page 74

REVIEW EXERCISES

1. What is the multiplicative inverse of ?

2. For what values of b is the rational expression undefined?

3. Write as an infinitely repeating decimal.

In 4–15, perform the indicated operations, express the answer in simplest form,and list the values of the variables for which the fractions are undefined.

4. 5.

6. 7.

8. 9.

10. 11.

12. 13.

14. 15.

In 16–19, simplify each complex rational expression and list the values of thevariables (if any) for which the fractions are undefined.

16. 17. 18. 19.

In 20–27, solve and check each equation or inequality.

20. 21.

22. 23.

24. 25.

26. 27.

28. The ratio of boys to girls in the school chorus was 4 : 5. After three moreboys joined the chorus, the ratio of boys to girls is 9 : 10. How many boysand how many girls are there now in the chorus?

29. Last week Stephanie spent $10.50 for cans of soda. This week, at the samecost per can, she bought three fewer cans and spent $8.40. How many cansof soda did she buy each week and what was the cost per can?

12x 1 1 2 2 # 81

x 1 3 . 7x

2d 1 13 5 2d 1 2

d 2 2xx 2 3 1 6

x(x 2 3) 5 5x 2 3

2xx 2 2 5 3x

x 1 13y 1 1

2 5 6y

x5 1 9 5 12 2 x

4a7 1 3

14 5 5a2

a 2 1 2 20a

1 2 16a2

x12 2 1

2x2

12 2 3

a 1 bb

1 1 ba

1 1 12

2 1 14

Ax 2 1x B 4 (1 2 x)A1 1 1

x 1 1 B ? A 1 1 3x2 2 4 B

A 1a 2 4 1 3

4 2 a B ? 16 2 a2

21

a 1 2 1 aa2 1 a 4 a

a 1 1

35b ? 5b 1 10

6 2 1b

d2 1 3d 2 184d 4 2d 1 12

8

yy 2 3 2

18y2 2 9

2a 1 1 1 3

a2 2 1

a2 1 2a5a 4 a2 1 7a 1 10

a23x 1 12

4x ? 2x2 2 16

25a 1 1

4a3

5ab2 ? 10a2b9

512

2(b 1 1)b(b2 2 1)

72

Review Exercises 75

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30. On a recent trip, Sarah traveled 80 miles at a constant rate of speed. Thenshe encountered road work and had to reduce her speed by 15 miles perhour for the last 30 miles of the trip. The trip took 2 hours. What wasSarah’s rate of speed for each part of the trip?

31. The areas of two rectangles are 208 square feet and 182 square feet.The length of the larger is 2 feet more than the length of the smaller. If the rectangles have equal widths, find the dimensions of each rectangle.

Use

Exploration

The early Egyptians wrote fractions as the sum of unit fractions (the reciprocalsof the counting numbers) with no reciprocal repeated. For example:

and

Of course the Egyptians used hieroglyphs instead of the numerals familiarto us today. Whole numbers were represented as follows.

� � ��

� � � � � � � � � � � � � � � � � � � � � � � � �1 2 3 4 5 6 7 8 9

To write unit fractions, the symbol was drawn over the whole numberhieroglyph. For example:

1. Show that for n � 0, .

2. Write and using Egyptian fractions.

3. Write each of the fractions as the sum of unit fractions. (Hint: Use theresults of Exercise 1.)

a. b. c. d. e. 1118

2324

712

710

25

23

34

1n 5 1

n 1 1 1 1n(n 1 1)

5 5120

13 and

10 100 1,000 10,000 100,000 1,000,000 10,000,000

or

23 5 1

3 1 13 5 1

3 1 14 1 1

1234 5 2

4 1 14 5 1

2 1 14

arealength 5 width. BA


14411C02.pgs 8/12/08 1:47 PM Page 76

CUMULATIVE REVIEW CHAPTERS 1–2

Part I

Answer all questions in this part. Each correct answer will receive 2 credits. Nopartial credit will be allowed.

1. Which of the following is not true in the set of integers?(1) Addition is commutative.(2) Every integer has an additive inverse.(3) Multiplication is distributive over addition.(4) Every non-zero integer has a multiplicative inverse.

2. 2 2 �3 2 5� is equal to(1) 0 (2) 2 (3) 6 (4) 4

3. The sum of 3a2 2 5a and a2 1 7a is(1) 3 1 2a (2) 3a2 1 2a (3) 4a2 1 2a (4) 6a2

4. In simplest form, 2x(x 1 5) 2 (7x 1 1) is equal to(1) 2x2 1 3x 1 1 (3) 2x2 1 12x 1 1(2) 2x2 1 3x 2 1 (4) 2x2 2 7x 1 4

5. The factors of 2x2 2 x 2 6 are(1) (2x 2 3)(x 1 2) (3) (2x 1 3)(x 2 2)(2) (2x 1 2)(x 2 3) (4) (2x 2 3)(x 2 2)

6. The roots of the equation x2 2 7x 1 10 5 0 are(1) 2 and 5 (3) 27 and 10(2) 22 and 25 (4) 25 and 2

7. For a � 0, 2, 22, the fraction is equal to

(1) (3)

(2) (4)

8. In simplest form, the quotient equals

(1) (3)

(2) (4)

9. For what values of a is the fraction undefined?

(1) 22, 25, 7 (2) 2, 5, 27 (3) 25, 7 (4) 5, 27

10. The solution set of the equation �2x 2 1� 5 7 is(1) {23, 4} (2) � (3) {23} (4) {4}

2a 1 4a2 2 2a 2 35

7a4(2b 2 1)

43

3b2

4(2b 2 1)234

b2b 2 1 4 3b

8b 2 4

2 1a 2 22 1

a 1 2

1a 2 2

1a 1 2

2a 2 1

a 2 4a

Cumulative Review 77

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Part II

Answer all questions in this part. Each correct answer will receive 2 credits.Clearly indicate the necessary steps, including appropriate formula substitu-tions, diagrams, graphs, charts, etc. For all questions in this part, a correct numer-ical answer with no work shown will receive only 1 credit.

11. Solve and check: �7 2 2x� � 3

12. Perform the indicated operations and write the answer in simplest form:

Part III


13. Write the following product in simplest form and list the values of x forwhich it is undefined:

14. The length of a rectangle is 12 meters longer than half the width. The areaof the rectangle is 90 square meters. Find the dimensions of the rectangle.

Part IV


15. Find the solution set and check: 2x2 2 5x � 7

16. Diego had traveled 30 miles at a uniform rate of speed when he encoun-tered construction and had to reduce his speed to one-third of his originalrate. He continued at this slower rate for 10 miles. If the total time forthese two parts of the trip was one hour, how fast did he travel at eachrate?

5x 1 5x2 2 1 ? x

2 2 x15

3a 1 5 2 a 2 3

5 4 a2 2 915


14411C02.pgs 8/12/08 1:47 PM Page 78

Documents

THE RATIONAL NUMBERSbase-ten system to fractions was given by Simon Stevin (1548–1620),who introduced to 16th-century Europe a method of writing decimal fractions. The decimal that